optimization of a box

the length of a cedar chest is twice its width. the cost/dm^2 of the lid is four times the cost dm^2 of the rest of the cedar chest. if the volume of the chest is 1440 dm^3 find the dimensions so that the cost is a minimum.

this is what i got so far

V=L x W x H
1440=2x(x)h
1440=2x^2(h)
h=720/x^2

C=4c(area of lid) + c(area of base) + c(area of sides)

That is what i got for cost and for the derivative, i'm having trouble solving for x and i should be able to get the dimensions once i have the value of x

You have too many constants in your derivative of the cost function. There should only be 1 2880 and 1 1440. Otherwise it seems right.

Set C' = 0 and solve for x. The 'c' cancels out (you could also have set it equal to 1 at the start to make life easier - your answer doesn't depend on the actual cost of the wood, just the relative cost).